Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Sum of three dice is eleven the sample space of one toss of three dice is:
$\Omega = \left \{ (1,1,1), ..., (6,6,6) \right \}$
so there are $6^3 = 216$ possible outcomes.
What is the probability to obtain an outcome where the sum of its three components is equal to 11?
I've considered the possible value can assume dice... | For the favorable cases: Since the dice can assume integer values between $1$ and $6$, we wish to find the number of solutions in the positive integers of the equation
$$x_1 + x_2 + x_3 = 11 \tag{1}$$
subject to the restrictions that $x_k \leq 6$ for $1 \leq k \le 3$.
A particular solution of equation 1 corresponds... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to solve $ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $ I am new to modulus and inequalities , I came across this problem:
$ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $
How to find $ x $ ?
| If $x\geq 0$, then
*
*$|x+1| = x+1$
*$|2^x-1| = 2^x-1$
So the equation becomes
$$2^{x+1}-2^x=2^x-1+1\\
2^{x+1} - 2\cdot 2^x=0\\
2^{x+1}-2^{x+1}=0\\
0=0$$
so it is always true.
There are two more cases to do: $-1\leq x<0$ and $x<-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many positive integer cubes are divisors of $1!\cdot 2! \cdot 3!\cdot 4! \cdot 5!\cdot 6!\cdot 7!\cdot 8!$? What I have tried:
$$N= 1!\cdot 2! \cdot 3!\cdot 4! \cdot 5!\cdot 6!\cdot 7!\cdot 8!\\
= 1^8\cdot 2^7\cdot 3^6\cdot 4^5\cdot 5^4\cdot 6^3\cdot 7^2\cdot 8^1 \\
= 2^{23}\cdot 3^9\cdot 5^4 \cdot 7^2$$
| You're off to a good start.
Hint 1: Let $n^3$ be a positive integers cube. Suppose $n$ has the prime factorization $n = p_1^{a_1} p_2^{a_2} \cdots p_n^{a_n}$. What is the prime factorization of $n^3$? In general, what can you say about the exponents of the prime factors of a perfect cube?
Hint 2: Let's simplify the pro... | {
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"source": "stackexchange",
"question_score": "1",
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In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digits must not be adjacent?
In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digit must not be adjacent?
I tried this by first taking total permutation as $\dfrac{8!}{2^4}$
Now $n_1$ as $22$ or $33$ or $44$ or $55$ oc... | You need a few more inclusion-exclusion steps to complete this approach.
Without constraints, you do indeed have $\dfrac {8!}{2^4} = 2520 $ arrangements.
Then there are $\dfrac {7!}{2^3} = 630$ cases where a $22$ is found in the arrangement, and similarly for the other digits.
Then there are $\dfrac {6!}{2^2} = 180$ ca... | {
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"source": "stackexchange",
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Solving system of differential equations to evaluate $x(t)$ I have to solve the following system of equations and find $x(t)$
$$\frac{dx}{dt} =a\cos(y)-b$$
$$x\frac{dy}{dt}=-a\sin(y)$$
Initial conditions:
*
*$t=0,x=x_o, y=\frac{\pi}{2}$
By dividing equations and integration, i obtained $x(y)$:
$$x=x_o \csc(y) \lef... | Too long for a comment.
Domains of my math competence are far from differential equations, so I can miss some subtleties. I assume that $a$ and $b$ are constants. Given a differentiable function $u=u(t)$ by $u’$ I shall denote its derivative with respect to $t$.
Differentiating the first equation we obtain
$x’’=-a\s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\si... | Let $\sin\theta=\sqrt{\frac37}$, so that $\cos\theta=\sqrt{\frac47}$ and $$\sin(\theta-30°)=\sqrt{\frac37}\sqrt{\frac34}-\sqrt{\frac47}\sqrt{\frac14}=\frac1{\sqrt{28}}.$$
Then by the triple angle formula
$$\sin(3\theta-90°)=\frac3{\sqrt{28}}-\frac4{28}\frac1{\sqrt{28}}=\frac{10}{7\sqrt 7}\approx0.54,$$
while
$$\sin(3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
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Question based on inequality with $\sum_{i=1}^{n}x_{i}=1$ Question:
If $n$ be given positive integer,let $x_{i}\ge 0$ such that $x_{1}+x_{2}+\cdots+x_{n}=1$. Find the maximum of the value $$x_{1}+x_{1}x_{2}+x_{1}x_{2}x_{3}+\cdots+x_{1}x_{2}x_{3}x_{4}\cdots x_{n}.$$
Try: (1):when $n=1$,it is clear equal $1$.
(2):when... | By AM-GM we obtain:
$$x_{1}+x_{1}x_{2}+x_{1}x_{2}x_{3}+\cdots+x_{1}x_{2}x_{3}x_{4}\cdots x_{n}$$
$$\leq x_1(1+x_2)\cdots(1+x_n)\leq\left(\frac{n-1+x_1+\cdots+x_n}{n}\right)^n=1.$$
The equality occurs for $x_2=\cdots=x_n=0$ and $x_1=1$, which says that $1$ is a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help understanding why converting repeating decimal to fraction works I recently learned how to convert a repeating decimal like $3.424242...$ to a fraction. I was however wondering why that actually works. After reading a few resources, I tried to understand it from a previous question on this site.
The explanation g... | In the equation $$10^m x = t + f$$ we have two "annoying" variables, $f$ and $t$ (annoying in the sense that they are not written as a fraction). The two variables are $f$ and $x$.
Our goal is to transform $x$ into a fraction, and since $t$ is already a fraction, the equation shows that we will be able to write $x$ as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
I... | Yes, we can give a simpler and more conceptual inductive proof. Notice that
$\qquad\qquad a(a^2+2)\, =\, a(a^2\!-\!1 + 3)\, =\, \color{#0a0}{(a-1)a(a+1)} + 3a$
so it suffices to show that one of any $\rm\color{#0a0}{3\ consecutive\ integers}$ is divisible by $3$
This has a simple inductive proof. Note that shifting s... | {
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Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a so... | $$\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}=\frac{r}{4R}\leq\frac{1}{8}.$$
Another way.
Since $f(x)=\ln\sin\frac{x}{2}$ is a concave function, by Jensen and AM-GM we obtain:
$$\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\leq\sin^2\frac{\frac{\alpha}{2}+\frac{\beta}{2}}{2}\sin\frac{\gamm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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How to calculate $\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$?
$$\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$$
I can't seem to find away to get rid of the $3^x$ and $4^x$ and then resolve it.
| Numerator : $7x^4 + x^2 3^x + 2 \lt 3x^2 3^x$
for $x\gt 4$ (say).
Denominator: $x^3 + x4^x +1 \gt x4^x$
for $x \gt 0.$
$0 \lt \dfrac{7x^4 + x^2 3^x +3}{x^3 +4^x +1} \lt $
$ \dfrac{3x^23^x}{x4^x} = 3x (\dfrac{3}{4})^x =
3x(\dfrac{ 4}{3})^{-x} $.
Note: $y := (\dfrac{4}{3})^{-x} = e^{-x \log(4/3)}$.
Let $a:= \log(4/3) \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 5
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An olympic mathematics problem regarding Cauchy-Schwarz This question was asked in Turkish National Maths Olympiad in 2008.
For all $xy=1$ we have $((x+y)^2+4)\cdot ((x+y)^2-2) \ge A\cdot(x-y)^2$.
What is the maximum value $A$ can get?
My efforts regarding this problem;
$(x+y)^2-8 \ge A\cdot(x-y)^2$
Using the propert... | Note that your first expression:
$$(x+y)^2-8\ge A\cdot (x-y)^2$$
must be:
$$(x+y)^4+2(x+y)^2-8\ge A\cdot (x-y)^2.$$
Alternative solution: You can also denote:
$$(x+y)^2=z; \ \ \ (x-y)^2=(x+y)^2-4xy=z-4.$$
Then:
$$(z+4)(z-2)\ge A\cdot (z-4) \Rightarrow$$
$$z^2+(2-A)z+4A-8\ge 0$$
This inequality is true for all $z$ (al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Upper Bound for $\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$ I was reading a solution for an analysis problem and they argued that $$\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$$ for $x\geq 1$ and $n\geq 1$, $x$ a real number and $n$ a natural number. Why is that?
| First the identity definitely doesn't hold for $n=1$ so let's assume $n\geq 2$, then we have
$$
\begin{align*}
\ln(1 + nx) &= \ln(nx) +\ln(1+1/nx) \\
&\leq \ln(nx) + 1/nx =\frac{ x^2 \ln (nx) + x/n }{x^2}.
\end{align*}
$$
Now
\begin{align*}
x^2\ln(nx) + x/n &= \left[ 1+x^2\ln(n) \right] \left( \frac{\ln(n)}{\ln(n) + x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the Coefficient of $(x^3 + 2y^2)^n$ containing $x^{18} y^{12}$ I was helping somebody when I scratched my head because of this question. It goes like
this:
If the middle term of the expansion of $(x^3 + 2y^2)^n$ is $C x^{18} y^{12},$ find C.
My work:
I let $u = x^3$ and $v = 2y^2.$ Then doing this: $$(u)^6 =... | The general term of the expansion $(x^3+2y^2)^n$ is
$$\binom{n}{r}2^{n-r}x^{3r}y^{2n-2r}$$
For the term of $x^{18}y^{12}$, take $r=6$ and such that $n=12$
So the coefficient, $C=\binom{12}{6}2^{12-6}=59136$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Solving a System of Two Differential Equations and getting the wrong answer Problem:
Solve the following system of differential equations.
\begin{eqnarray*}
5x' + y' - 3x + y &=& 0 \\
4x' + y' - 3x &=& -3t \\
\end{eqnarray*}
Answer:
To solve this, we use the operator method.
\begin{eqnarray*}
(5D-3)x + (D+1)y &=& 0 \\
... | from the second equation we obtain:
$$y'=-3t+3x-4x'$$ plugging this in the first equation:
$$x'-y=3t$$ differentiating this with respect to $t$:
$$x''-y'=3$$ and again we have
$$x''-(-3t+3x-4x')=3$$ this is easy to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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An isosceles triangle with sides... An isosceles triangle with sides $|AB|=6$, $|BC|=|AC|=5$. Calculate the height of the triangle, against the side $BC$.
I drew the following:
By the figure I get that $|CD|=\sqrt{25-x^2}$ and $|BD|=5-\sqrt{25-x^2}.$ Using the pythagorean theorem on $\triangle ABD,$ I get $$36=x^2+\le... | An easy way to solve this problem is to compute the area of the triangle using Heron's formula. In particular, $s = (6+5+5)/2=8$, and $A = \sqrt{8(8-6)(8-5)^2}=12$. But $A=\frac{5x}{2}$. Therefore, $x = \frac{24}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or ... | Hint:
$
5^{2n+1} 2^{n+2} + 3^{n+2} 2^{2n+1}
\\= 20\cdot 50^n + 18 \cdot 12^n
\\= 19(50^n + 12^n) + 50^n - 12^n
\\= 19(50^n + 12^n) + (2\cdot 19 +12)^n - 12^n
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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About using the lagrange multiplier. Today I have seen a question like the following;
$x+y+z=5$ and $xy+yz+xz=3$ and $x,y,z \in \mathbb{R}^+$
What is the maximum value $x$ can get?
Now it is pretty obvious that question is solvable using many simple elementary methods like say $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25$
$25... | We will use the quadratic equation approach to determine the min and max values of $x$. We have: $y+z = 5-x$, and $yz + x(y+z) = 3\implies 3- x(5-x)= yz\implies x^2-5x+3= yz\le \dfrac{(y+z)^2}{4}= \dfrac{(5-x)^2}{4}\implies 4x^2-20x+12 \le 25-10x+x^2\implies 3x^2-10x-13 \le 0 \implies (3x -13)(x + 1)\le 0\implies -1 \l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$
Olympiad Inequation
Let $x$, $y$ and $c$ be distinct real numbers. Prove that:
$$\left(\frac{2x-y}{x-y}\right)^2+\left(\frac{2y-z}{y-z}\right)^2+\left(\frac{2z-x}{z-x}\right)^2 \g... | We want to minimize
$$
\begin{align}
&\underbrace{\left(\frac{mx-ny}{x-y}\right)^2}_{\left(\frac{m+n}2+\frac{m-n}2\frac{x+y}{x-y}\right)^2}+\underbrace{\left(\frac{my-nz}{y-z}\right)^2}_{\left(\frac{m+n}2+\frac{m-n}2\frac{y+z}{y-z}\right)^2}+\underbrace{\left(\frac{mz-nx}{z-x}\right)^2}_{\left(\frac{m+n}2+\frac{m-n}2\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
I have reached:
$x^2(y-1) + x(y-3) + (y-1) = 0$
I also know that the denominator of $y= f(x)$ is greater than $0$.
How do I continue from here? I am unable to find a suita... | $$y=1+\frac{2x}{x^2+x+1}$$
The function $y(x)$ is continuous and differentiable in $\Bbb R$. Moreover
$\lim_{x\to\pm\infty}y(x)=1$, so $y(x)$ is bounded. This means that the range of $y$ is an interval $I$.
Then
$$y'=\frac{2x^2+2x+2-4x^2-2x}{(x^2+x+1)^2}=\frac{2-2x^2}{(x^2+x+1)^2}$$
which vanishes at $x=\pm1$.
Compute ... | {
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"url": "https://math.stackexchange.com/questions/2420197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Logarithmic problem If $\log_{16} 15 =a$ and $\log_{12} 18 =b$, then show that $$\log_{25} 24 = \frac{5-b}{16a-8ab-4b+2}.$$
| Let $x=\log _{25}24$ so that $25^x=24$
From $16^a=15$ we get $$2^{8a x}=16^{2a x}=15^{2 x}= 3^{2 x}5^{2x}=3^{2x}24=2^33^{2x+1}$$
$$2^{8a x-3}=3^{2x+1}$$
From $12^b=18$ we get
$$2^{2b-1}=3^{2-b}$$
Multiplying these two equations we have
$$(8a x-3)(2-b)=(2b-1)(2x+1)$$
Solve for $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Proof verification: Determine whether $f(x)$ is one-to-one or not
$$f:\mathbb{R}\mapsto\mathbb{R}
\text{ defined as } f(x)=2x^3+3x-4 \\
\underline{\textit{proof(by contradiction)}}\\
\text{By definition a function is called one-to-one if}\\
f(x)=f(y),\text{ } x,y\in\mathbb{R} \text{ implies that x=y.}\\
\text{Suppose... | $$f(x) = f(y)$$
$$\Rightarrow2x^3 + 3x - 4 = 2y^3 + 3y - 4$$
$$\Rightarrow 2(x^3-y^3) + 3(x-y) = 0$$
$$\Rightarrow 2(x-y)(x^2+xy+y^2) + 3(x-y) = 0$$
$$\Rightarrow (x-y)(2x^2 + 2xy + 2y^2 +3) = 0$$
$$\Rightarrow x = y$$
The last step can be justified by saying that the right factor is never $0$. I leave this as an exe... | {
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"url": "https://math.stackexchange.com/questions/2423110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\max(ab+bc+ac)$
Let a,b,c be real numbers such that $a+2b+c=4$. What is the value of $\max(ab+bc+ac)$
My attempt:
Squaring both the sides:
$a^2 +4b^2+c^2+2ac+4bc+4ab=16$
Then I tried factoring after bringing 16 to LHS but couldn't. It's not even a quadratic in one variable or else I could have directly f... | From the given condition $\,2b=4-a-c\,$, then:
$$\require{cancel}
\begin{align}
2(ab+bc+ca) &= 2b(a+c)+2ac \\
&= (a+c)(4-a-c)+2ac= \\
&= 4a - a^2 - \bcancel{ac} + 4c - \bcancel{ac} - c^2 + \bcancel{2ac} = \\
&= \color{red}{8} -a^2 + 4a \color{red}{-4} - c^2 +4c \color{red}{-4} = \\
&= 8 - (a-2)^2 - (c-2)^2 \\
&\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ende... | $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}
=(a-b)\lim_{x\rightarrow \infty}\frac{x}{\sqrt{x^2 +ax} - \sqrt{x^2 +bx}}=(a-b)\lim_{x\rightarrow \infty}\frac{x}{\vert{x}\vert\left(\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}\righ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 5,
"answer_id": 0
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$66$ points in $100$ shots. I just received a probability problem from a friend via a text and by the time I took to read it, I was sent a solution as well - which is confusing.. The question goes like this..
A person shoots basketball 100 times. First time he scores a point and second time he misses it. For the foll... | Let $P(x, n)$ be the probability that you have $x$ points after $n$ throws. We are given that $$P(1,2) = 1$$
You are also given information about the probability of scoring at any step, if you know the number of points previously scored. In particular:
$$P(\text{scoring from } x-1, n-1 ) = \frac{x-1}{n-1}$$
$$P(\text{s... | {
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"source": "stackexchange",
"question_score": "17",
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"answer_id": 2
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Is writing equalities within an equation abuse of notation? I'll occasionally write equalities within parentheses or sqrt signs to make my steps more compact.
E.g.:
$$ r = \frac{3}{4}\sqrt[4]{\frac{7}{9}\cdot\frac{16}{7}=\frac{16}{9}} = \frac{3}{4}\sqrt{\frac{4}{3}}=\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} $$
I always ... | Your notation is very non-standard and will be confusing to most people. Many people use curly braces to denote intermediate results:
$$
r = \frac{3}{4}\underbrace{\sqrt[4]{\frac{7}{9}\cdot\frac{16}{7}}}_{\sqrt[4]{\frac{16}{9}}} = \frac{3}{4}\sqrt{\frac{4}{3}}=\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}
$$
| {
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"source": "stackexchange",
"question_score": "3",
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Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't unde... | For example, if $n$ is odd, $n=2k-1$ then we obtain:
$$LS=\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{k(k-1)(2k-1)}{6}=$$
$$=\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{\frac{n+1}{2}\cdot\frac{n-1}{2}\cdot n}{6}=\frac{n(n+1)}{2}$$
| {
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How to compute $S_{2016}=\sum\limits_{k=1}^{2016}\left(\sum\limits_{n=k}^{2016}\frac1n\right)^2+\sum\limits_{k=1}^{2016}\frac1k$? I came across a question asking the value of the following sum:
\begin{align}
\left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+\left(\frac{1}{2}+\frac{1}{3... | (new solution - a bit shorter)
$$\require{cancel}\begin{align}
&\;\;\;\sum_{j=1}^n\left(\sum_{r=j}^n\frac 1r\right)^2+\sum_{j=1}^n\frac 1j\\
&=\sum_{j=1}^n\left(\sum_{r=j}^n\sum_{s=j}^n\frac 1{rs}\right)+\sum_{j=1}^n\frac 1j\\
&=\sum_{j=1}^n\left[2\sum_{r=j}^n\sum_{s=r}^n\frac 1{rs}-\sum_{r=j}^n\frac 1{r^2}\right]+\sum... | {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
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Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$... | \begin{eqnarray*}
1^2+3^2+5^2+\cdots+(2n-1)^2=\frac{(2n-1)n(2n+1)}{3}
\end{eqnarray*}
So
\begin{eqnarray*}
1^2+3^2+5^2+\cdots+(2n-1)^2 + (2n+1)^2 &=& \frac{(2n-1)n(2n+1)}{3} + (2n+1)^2 \\
&=& \frac{(2n+1)}{3} \underbrace{\left( n(2n-1) +3(2n+1) \right)}_{2n^2+5n+3} \\
&=& \frac{(2n+1)(n+1)(2n+3)}{3}.
\end{eqnarray*... | {
"language": "en",
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"source": "stackexchange",
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Find minimum possible values of the leading coefficient of quadratic equation let the polynomial $$f(x) = ax^2- bx+ c$$ (where a, b, are positive integers). If $f(p)=f(q)=0$ where $0<p<q<1$ then what is the minimum possible value of a.
I used some basic inequalities to get the following -
$$1) b<2a$$
$$2) a>c$$
$$3) a... | You are on the right track. First of all, you don't really need condition $1)$ since it is implied by $2)$ and $3)$. On the other hand, you need to add condition $b^2>4ac$, for example $2x^2-x+1$ satisfies your conditions but has no real roots.
Now, you can exploit the fact that $a,b,c$ are positive integers to get bet... | {
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"source": "stackexchange",
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$(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$. Prove that at least one from $a,b,c$ is 2.
Assume that $\{a,b,c\} \subset \Bbb R$, $(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$.
Prove that at least one of the numbers $a, b, c\ $ is 2.
This is from a list of problems used for training a team for a m... | Suppose that $a=2$. Then $b+c=0$ and hence $b=-c$. The third equation gives $b^2+c^2=2$ hence $2b^2=2$ and $b^2=1$. If $b=1$ then $c=-1$ and vice versa. Then $(2-3)^3+(1-2)^3+(-1-2)^3=-1-1-27=-29$. The same thing happens when $b=-1$ and $c=1$.
If $b=2$, then $a=-c$ again which leaves with $a$ being either $1$ or $-1$.... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Probability about oranges There are $15$ oranges in each bag.
Each orange has a probability of $10$% to be bad-looking.
Supermarkets only sell the bags that have $12$ or more good-looking oranges.
What percentage of all the bags will be sold?
| Compute:
$$
\ \ \ \
{15 \choose 3} \cdot Pr(3 \ \text{bad-look oranges})
\cdot Pr(12 \ \text{good-look oranges})
\\ +
{15 \choose 2} \cdot Pr(2 \ \text{bad-look oranges})
\cdot Pr(13 \ \text{good-look oranges}) \\ +
{15 \choose 1} \cdot Pr(1 \ \text{bad-look oranges})
\cdot Pr(14 \ \text{good-look oranges}) \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Induction proof of $F(n)^2+F(n+1)^2=F(2n+1)$, where $F(n)$ is the $n$th Fibonacci number.
Let $F(n)$ denotes the $n$th number in Fibonacci sequence. Then for all $n\in\mathbb{N}$,
$$F(n)^2+F(n+1)^2=F(2n+1).$$
I know how to prove it by using the formula
$$F(n)=\frac{\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-... | \begin{align} &F(x)=F(x-1)+F(x-2) = F(2)F(x-1)+F(1)F(x-2). \ \\ \ \\ &F(x-1)=F(x-2)+F(x-3) \\ &\Rightarrow F(x)=F(2)(F(x-2)+F(x-3))+F(1)F(x-2)=(F(1)+F(2))F(x-2) \\ &+F(2)F(x-3)=F(3)F(x-2)+F(2)F(x-3). \ \\ \ \\ &F(x-2)=F(x-3)+F(x-4). \\ &\Rightarrow F(x)=F(3)(F(x-3)+F(x-4))+F(2)F(x-3)=(F(2)+F(3))F(x-3) \\ &+F(3)F(x-4)=F... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Determine for what values of $x$ the given series converges The given series is
$\sum_{n=1}^∞
(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n}
)x^{n} $.
I tried it by using Cauchy Root Test as follows-
Let
$y=\lim_{n\to\infty}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{... | There are two serious problems. The first one is your assertion that
$$
\log\Bigl(1+\frac12+\dots+\frac1n\Bigr)=-\infty.
$$
It is well known (harmonic series) that
$$
\lim_{n\to\infty}\Bigl(1+\frac12+\dots+\frac1n\Bigr)=+\infty,
$$
so that
$$
\lim_{n\to\infty}\log\Bigl(1+\frac12+\dots+\frac1n\Bigr)=+\infty.
$$
The seco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\lim\limits_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$ This problem
$$\lim_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$$
is from Silverman's "Modern Calculus and Analytical Geometry" Section 22, #16d. I've been struggling on it for a while a... | Let
$$a=(x+1)^{\frac{1}{6}}$$
$$b=(1-x)^{\frac{1}{6}}$$
Our equation becomes,
$$\lim_{x\to 0} \frac{a^3-b^3}{a^2-b^2}$$
$$\lim_{x\to 0} \frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}$$
$$\lim_{x\to 0} \frac{a^2+ab+b^2}{a+b}$$$$\lim_{x\to 0} \frac{(x+1)^{\frac{1}{3}}+(1-x^2)^{\frac{1}{6}}+(1-x)^{\frac{1}{3}}}{(x+1)^{\frac{1}{6}}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Binomial expansion of complex variable For the binomial expansion $ (1+x)^n = p_0 + p_1x+p_2x^2+...,$ where
$p_i$ refer to the binomial coefficients.
Need to substitute the cube roots of unity (1, w, $w^2$) for the variable x; with w = $\dfrac{-1+i\sqrt{3}}{ 2},$
find the sums:\begin{align} (a) p_0 + p_3 + p_6 + ...... | Ill do part 2. Since all coefficients of $x^{1+3k}$ are to be retained, we multiply by $x^2$ to make the power $x^{3+3k}$, ie a multiple of $3$.
$$(1+x)^n = a_o + a_1 x + a_2 x^2...a_nx^n$$
$$x^2(1+x)^n = a_ox^2+a_1x^3+a_2 x^4 ... a_nx^{n+2}$$
Now substituting cube roots of unity in above equation leaves required term... | {
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Find the area of a triangle using the sides lengths
The sides lengths of a triangle $a,$ $b$ and $c$ verify: $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=12b.$$ The task is to find the area of the triangle.
I'm trying to apply the heron's formula: $$\dfrac{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}} {4}.$$
How do i get to heron... | $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144-12b=\sqrt{a-24} + (b-6)^2 +|c-12\sqrt{3} |+108>0.$$
Thus, we have no any triangle with your given.
If you mean $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=24b$$ then we get $$(a,b,c)=(24,12,12\sqrt3)$$ and now you can use the Heron's formula.
| {
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Show that $\limsup \frac{\epsilon_1+\cdots +\epsilon_n}{n}\geq 0$？ Suppose $a_n$ is a sequence such that $a_n \downarrow 0$,
$\epsilon_n=-1$ or $1$ for all $n$, the series $\sum a_n$ diverges but the series $\sum_{n=1}^\infty \epsilon_n a_n$ converges. Is it true that $\limsup \frac{\epsilon_1+\cdots +\epsilon_n}{n}\... | The statement is not true. Using Michael's construction, but simplified gives:
$$\epsilon_n = \{1, -1, -1, 1, 1, 1,1, \dots\}$$
$$\epsilon_n = (-1)^{\lfloor \log_2 n\rfloor}$$
After $1$, $7$, $31$, $\dots$, $2^{2k+1} - 1$ numbers we hit a maximum. This maximum is exactly:
$$\sum_{i=0}^{2k} (-2)^i = \frac{1}{3}(2^{2k+1}... | {
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"source": "stackexchange",
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Why is $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3$?
How to show that $$\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3?$$
This equality comes from solving $$t^3 - 15 t - 4 = 0$$ using Cardanos fomula and knowing the solution $t_1=4$.
I have attempted multiplying the whole thing with $(\sqrt[3]{18+5\... | Let $(a + b\sqrt{13})^3 = (18 + 5\sqrt{13})$ for $a, b \in \Bbb Q$
Expanding the LHS gives,
$$(a^3 + 39 ab^2 - 18 ) +\sqrt{13}(3a^2 b + 13 b^3 - 5) = 0$$,
From this we get,
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
Solving the system give $ a = \dfrac 32$ and $ b = \dfrac12$
Therefor... | {
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"question_score": "6",
"answer_count": 4,
"answer_id": 1
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If $u_{n+1} = u_n^2 - u_n + 1$ and $u_0=a>1$, show that $\sum\frac{1}{u_n} \rightarrow \frac{1}{a-1}$ If $(u_n)\in\mathbb{R}^\mathbb{N},u_{n+1} = u_n^2 - u_n + 1$ and $u_0=a>1$, show that $\sum\dfrac{1}{u_n} \rightarrow \dfrac{1}{a-1}$.
I don't even know how to prove that the series converges
| The key observation, which kills theproblem quite instantly, is the following: for every $n$ it holds the relation $$\frac{1}{u_n} = \frac{1}{u_n - 1} - \frac{1}{u_{n + 1} - 1}$$ Indeed we have $$\frac{1}{u_n - 1} - \frac{1}{u_{n + 1} - 1} = \frac{1}{u_n - 1} - \frac{1}{u_n(u_n - 1)} = \frac{u_n - 1}{u_n(u_n - 1)} = \f... | {
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Find the number of solutions of x+y+z=17? Find the number of solutions of $x+y+z=17$ where $2\le x\le 5, 3\le y \le 6, 4\le z\le7$.
My approach:
The number of solutions with the indicated constraints is the coefficient of $x^{17}$ in the expansion of
($x^2+x^3+x^4+x^5)(x^3+x^4+x^5+x^6)(x^4+x^5+x^6+x^7)$
I have changed... | I double counted some solutions. The 3 solutions which I added to 45 to get 48 must be subtracted from 45 to get 42. This is because these 3 solutions were counted twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the matrix of a linear transformation If T : $\mathbb R^{3}$$\mapsto$$\mathbb R^{3}$ is a linear transformation such that
T $\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}$ = $\begin{pmatrix}
3 \\
1 \\
4 \\
\end{pmatrix}$, $T$ $\begin{pmatrix}
... | To expand on my comment above since @Gregory already gave the answer using one method:
Given information on how $T$ acts on the standard basis, knowing $T\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}3\\1\\4\end{bmatrix}$ and so on, this tells you exactly how to represent $T$ as a matrix. The columns of $T$ are q... | {
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"source": "stackexchange",
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Determine the limit of a sequence using squeeze theorem $$ \left(\frac{\left(1+2\left(-1\right)^x\left(x+4x^2\right)\right)}
{5+7x^3}\right)\left(\sin \left(\frac{\left(3x\pi \right)}{7}\right)\ +\cos
\left(\frac{1}{x}\right)\right)
$$
I have to find the limit of this equation. I am confused by the (-1)^x and... | Noting
$$ \bigg|\left(\frac{\left(1+2\left(-1\right)^x\left(x+4x^2\right)\right)}
{5+7x^3}\right)\left(\sin \left(\frac{\left(3x\pi \right)}{7}\right)\ +\cos
\left(\frac{1}{x}\right)\right)\bigg|\le\frac{2\left(1+2\left(x+4x^2\right)\right)}
{5+7x^3}
$$
and
$$ \lim_{x\to\infty}\frac{2\left(1+2\left(x+4x^... | {
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The manager chooses 2 guests at random. Find the probability that they are both staying in the same room. There are $25$ rooms in a hotel; $3$ with $0$ guests, $4$ with $1$ guest each, $12$ with $2$ guests each, $5$ with $3$ guests each and $1$ with $4$ guests.
Total rooms: $25$
Total guests: $47$
Q. The manager choose... | There are
$$\binom{47}{2}$$
ways for the manager to select two of the forty-seven guests in the hotel.
In order to select two guests from the same room, the manager must select both guests from a room with two people or two of the three guests in a room with three people or two of the four guests in a room with four p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $c^{a} + c^{b} = a^{c} + b^{c}$ when $a \ne b$? Right now I'm working on a computer algorithm that manipulates a binary tree data structure. Within this tree, there are many "gaps"; i.e., nodes that don't produce children nodes, thus resulting in gaps that become increasingly large as the tree grows.
That being sa... | The only triples of positive integers $(a, b, c)$ for which $a \neq b$ and $c^a + c^b = a^c + b^c$, up to interchange of $a$ and $b$, are $(1, 3, 2)$, $(2, 4, 2)$, and $(2, 4, 4)$. Allowing $a = b$ adds the the solutions $(4, 4, 2)$ and $(2, 2, 4)$ as well as the trivial solutions $a = b = c$.
For $c$ fixed, $a^c + b^c... | {
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"timestamp": "2023-03-29T00:00:00",
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An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because o... | Hint: The difference of two squares $a^2-b^2$ can be factored as $(a-b)(a+b)$. A sum of two squares $a^2+b^2$ can be thought of as $a^2-(ib)^2$...
| {
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"timestamp": "2023-03-29T00:00:00",
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In how many ways can a student score exactly $100$ points on four $50$ point exams? greater than $100$? A student takes up $4$ exams with $50$ points each. In how many ways can he score exactly $100$? Similarly, in how many ways can he score greater than $100$?
My try:
I cannot seem to fill in the $4$ possibilities. Th... | The scores he can get are the coefficients of the different powers of $x$ in the expansion
\begin{align*}
(1+x+\cdots + x^{50})^4 &= \left(\frac{1-x^{51}}{1-x}\right)^4 \\
&= (1-4x^{51} + 6X^{102} - 4X^{153} + X^{204})\left(1+ 4x + \frac{4\cdot 5}{1\cdot 2}x^2 + \frac{4\cdot 5 \cdot 6}{3!}x^3 + \cdots +\binom{n+3}{3}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Rationalize a fraction. Rationalize the denominator $\frac{1}{2^{\frac{1}{3}} + 3^{\frac{1}{3}} + 4^{\frac{1}{3}}}$. Is there a short solution for this task ? Thanks in advance.
| First, use the identity
$$(a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \;\; = \;\; a^3 + b^3 + c^3 - 3abc$$
with $a = \sqrt[3]{2}$ and $b = \sqrt[3]{3}$ and $c = \sqrt[3]{4}.$
Multiplying both the numerator and denominator of your fraction by the numerical form of $a^2 + b^2 + c^2 - ab - ac - bc$ will give you
$$ \frac{\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area of a triangle inscribed in the ellipse (Unicamp Mathematics Olympiad) Let $\zeta$ denote the Carteasian region given by $\zeta = \{(x,y): \frac{x^2}{4} + \frac{y^2}{9} = 1\}$. Let $A = (0,3), B = (x,y), C = (z,w)$ be points such that $A,B,C \in \zeta%$ and $\Delta ABC$ is equilateral. What is the area of ... | $A=(0,3)$ is a vertex of the given ellipse. If $ABC$ is equilateral, by symmetry we have that $B$ and $C$ share the same $y$-coordinate, hence they are points of the form
$$ B=\left(-2\sqrt{1-\frac{y^2}{9}},y\right),\qquad C=\left(2\sqrt{1-\frac{y^2}{9}},y\right) $$
with $y\in(-3,3)$. In order that $ABC$ really is equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$. I'm trying to find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$.
I can find the equation for the length pretty easily but I'm... | $$\int \sqrt{1+(x'(y))^2}\,dy=\int \sqrt{1+\left(\frac{y^{5/2}}{2}-\frac{1}{2 y^{5/2}}\right)^2}\,dy=\int\sqrt{\frac{1+\left(y^5-1\right)^2}{4 y^5}}\,dy=$$
$$=\int\frac{1}{2} \sqrt{\frac{\left(y^5+1\right)^2}{y^5}}\,dy=\frac12\int\frac{y^5+1}{\sqrt{y^5}}\,dy=\frac12\int\left(y^{-5/2}+y^{5/2}\right)\,dy=\frac{y^{7/2}}{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$? I think that the best way to prove this would be to prove by contradiction. Am I right?
If so, can I just assume that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ and say $a = 1$ and $b = 1$ and show that $\sqrt{1+1} \neq \sqrt{1} ... | Let $a,b \in \mathbb R^+$ and suppose for the sake of contradiction that $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$. Squaring both sides we get $$a+b = a+b+2\sqrt{ab}$$
$$\implies 2\sqrt{ab}=0 \implies ab=0$$ a contradiction as $a,b \neq 0$. Therefore $\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}$ anytime $a$ and $b$ are nonzero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving the ODE: $\frac{1}{r} \frac{d}{d r} \left( r \frac{d f}{d r} \right) + \left( a - b e^{r^2} \right) f = c+ d e^{- r^2} $ I'm trying hard to solve this:
$$\frac{1}{r} \frac{d}{d r} \left( r \frac{d f}{d r} \right) + \left( a - b e^{r^2} \right) f = c+ d e^{- r^2} $$
where $r$ ranges between $0$ and $\infty$, $a... | Hint:
$\dfrac{1}{r}\dfrac{d}{dr}\left(r\dfrac{df}{dr}\right)+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
$\dfrac{d^2f}{dr^2}+\dfrac{1}{r}\dfrac{df}{dr}+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
Let $s=r^2$ ,
Then $\dfrac{df}{dr}=\dfrac{df}{ds}\dfrac{ds}{dr}=2r\dfrac{df}{ds}$
$\dfrac{d^2f}{dr^2}=\dfrac{d}{dr}\left(2r\dfrac{df}{ds... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Two-variable limit of $\lim_{(x,y)\to(0,0)}\frac{\sin(x^4+y^4)}{x^2+y^2}$ $$\lim_{(x,y)\to(0,0)}\frac{\sin(x^4+y^4)}{x^2+y^2}$$
I tried to bound it with $\frac{\sin((x^2+y^2)^2)}{x^2+y^2}$ and using polar coordinates with $x = r\cos\theta$ and $y = r\sin\theta$, but neither of the approaches provided any results. I kno... | Use $|\sin t|\leq |t|$ then
$$\Big|\frac{\sin(x^4+y^4)}{x^2+y^2}\Big|\leq\frac{x^4+y^4}{x^2+y^2}\leq\frac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c > 0$$
I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with d... | Consider you that you look for the minimum of function $$\Phi=\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \tag 1$$ Compute the partial derivatives
$$\frac{\partial\Phi}{da}=-\frac{b+c}{a^2}+\frac{1}{b}+\frac{1}{c}\tag 2$$
$$\frac{\partial\Phi}{db}=-\frac{a+c}{b^2}+\frac{1}{a}+\frac{1}{c}\tag 3$$
$$\frac{\partial\Phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Improper integral depending on parameter Find the value of the constant $C$ for which the integral $$\int \limits_{0}^{\infty}\left (\dfrac{1}{\sqrt{x^2+4}}-\dfrac{C}{x+2}\right)dx$$ converges. Evaluate the integral for this value of $C$.
I have some difficulties with above problem. I know some methods such as $x+2 \si... | First we have,
$\int_0^\infty \frac{1}{\sqrt{x^+4}}-\frac{C}{x+2}$
$=\int_0^\infty \frac{x+2-C\sqrt{x^2+4}}{(x+2)(\sqrt{x^2+4}}$
$=\int_0^\infty \frac{x+2-Cx\sqrt{1+\frac{4}{x^2}}}{(x^2+2x)(\sqrt{1+\frac{4}{x^2}}})$
$=\int_0^\infty \frac{x(1-C\sqrt{1+\frac{4}{x^2}})+1}{x^2\sqrt{1+\frac{4}{x^2}}+x\sqrt{1+\frac{4}{x^2}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Group with $[a,b][c,d] \ne [f,g]$ I guess that multiplication of commutators isn't equal to commutator of some elements. But I don't know how to find such group. Because my task became so : I tried to get four elements (for example in $GL_{2}$) , consider their product of commutators and then I stuck because it become ... | In general, $[G,G] \not = \{[g,h] :g,h \in G\}$. We shall illustrate this via $\operatorname{SL}(2,\mathbb{R})$, the group of real $2 \times 2$ matrices with determinant 1.
First we show $-I$ is not a commutator. Suppose, to the contrary, that $A^{-1}B^{-1}AB = -I$ for some $A,B \in SL(2,\mathbb{R})$. Then considering ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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is there a name for a series wich alternates + and - terms? Is there a specific name for a geometric series such as this?
$1-\frac{c (os\theta)}{k}+\frac{c^2}{k^2}-\frac{c^3}{k^3}+\frac{c^4}{k^4}-....$
How can we identify it if positive terms are even or odd?
Also, does its definition change if the numerator follows a... | It's the Taylor expansion of $\frac 1 {1+ \frac c k}$
$$1-\frac{c}{k}+\frac{c^2}{k^2}-\frac{c^3}{k^3}+\frac{c^4}{k^4}-....=\sum_{n=0}^{\infty}(-1)^n( \frac c k)^n=\sum_{n=0}^{\infty}( \frac {-c} k)^n=\frac 1 {1+ \frac c k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Limit of a function using Taylor series
How can I verify that
$\lim_{x \to 0} \dfrac{\cos{2x} - \sqrt{1-4x^2}}{2x \sin x^3} = \frac{4}{3}$?
I've tried with Taylor so that:
$2x\sin x^3 \sim 4x^4 - \frac{2}{3}x^{10}$
$\cos2x \sim 1- 2x^2$
$\sqrt{1-4x^2} \sim 1-2x^2-2x^4$
But it keeps me giving the wrong result, so ... | Hint. First of all $2x\sin(x)^3=2x^4+o(x^4)$. Moroever, here you need a longer expansion for $\cos(2x)$:
$$\cos(2x)=1-\frac{(2x)^2}{2}+\frac{(2x)^4}{4!}+o(x^4)=1-2x^2+\frac{2x^4}{3}+o(x^4).$$
Hence
$$\frac{\cos(2x) - \sqrt{1-4x^2}}{2x \sin x^3}=
\frac{1-2x^2+\frac{2x^4}{3} - (1-2x^2-2x^4)+o(x^4)}{2x^4+o(x^4)}
$$
Do you... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Lagrange multiplier to function $x^2+y^2+z^2$ Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given condition:
$$f(x,y,z)=x^2+y^2+z^2; \quad x^4+y^4+z^4=1$$
My solution: As we do in Lagrange multipliers I have considered $\nabla f=\lambda \nabla g$ where $g(x,y,z)=x^4+y^4+... | $x^2+y^2+z^2 \ge x^4+y^4+z^4 = 1$, and equality occurs when $x = 0,y = 0, z = \pm 1$ or permutations of them. Also by Cauchy-Schwarz inequality: $x^2+y^2+z^2 \le \sqrt{3(x^4+y^4+z^4)} = \sqrt{3}$ with equality occurs when $x = y = z = \pm \dfrac{1}{\sqrt[4]{3}}$ . Thus we can conclude that the min $= 1$ and the max $= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $\frac{1}{2} + \frac{1}{2(u+1)^2} - \frac{1}{\sqrt{1+2u}} \geq 0$ for $u \geq 0$ This inequality provides a tight lower bound to $\sqrt{1+2u}$ for $u\geq 0$ without a radical. I was trying to solve it by squaring the radical and cross-multiplying and repeated differentiation of the resulting expression, I wonder ... | It's
$$\sqrt{2u+1}(u^2+2u+2)\geq2(u+1)^2$$ or $$(2u+1)(u^2+2u+2)^2\geq4(u+1)^4$$ or
$$u^3(2u^2+5u+4)\geq0.$$
Done!
Also, we can use the following way.
Let $\sqrt{2u+1}=x$.
Thus, $x\geq1$ and we need to prove that
$$\frac{1}{2}+\frac{1}{2\left(\frac{x^2-1}{2}+1\right)^2}\geq\frac{1}{x}$$ or
$$1+\frac{4}{(x^2+1)^2}\geq\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Crazy Functional Equation - Possibly related to trig identities? $p(2x)=p(x)p(-x)-p(x-1)p(1-x)+4x-1$ I'm trying to find a set of nontrivial (identity function is not allowed, and neither are constant functions) continuous solutions to the functional equation
$$p(2x)=p(x)p(-x)-p(x-1)p(1-x)+4x-1$$
...and, to be honest, I... | One can take trial functions, say
$$p_{n}(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + \cdots + a_{n} \, x^n$$
and attempt to find the values of the coefficients.
As an example consider the case of $n=2$, or $p_{2}(x) = a_{0} + a_{1} \, x + a_{2} \, x^2.$ By expanding terms one will find that
\begin{align}
a_{0} + 2 a_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomial sum: $x^n+x^{n-1}y+x^{n-2}y^2+x^{n-3}y^3+\dots+xy^{n-1}+y^n$ Find sum of the expression,
$$x^n+x^{n-1}y+x^{n-2}y^2+x^{n-3}y^3+\dots+xy^{n-1}+y^n$$
where $x,y$ are real numbers and $n$ is a natural number.
| Hint: It’s a geometric progression with common ratio $\dfrac{y}x$.
| {
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Prove the existence of complex matrix that satisfies the following conditions. B is a 2 by 2 complex matrix. $B=0$ or $B^2 \neq 0$. Prove that a complex matrix C exists such that $C^2=B$.
I cannot understand what this problem wants me to do. "$B=0$ or $B^2 \neq 0$". Should I deal with the two cases? It would be great i... | Ok, if $B=0$ then take $C=0$ and you are done, so $B^2 \neq 0$.
Since $B$ is complex matrix it has Jordan normal form. There exists invertable matrix $S$ such that $B=SXS^{-1}$ where X has one of 2 possible forms:
$X=\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} $ or $X=\begin{pmatrix} \lambda & 1 \\ 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expansion of Binomial Coefficient $(1+x+x^2)^n$ Suppose $n$ is a natural number and consider expansion:
$$\left(1+x+x^2\right)^n=\sum_{r=0}^{2n} \ a_r x^r$$
Find $\ a_0+ \ a_3+ \ a_6+ \ a_9\ldots$
I used different method, but could not arrive at the answer.
| Note: I think the nice comment from @labbhattacharjee is worth an answer by its own. In order to obtain a formula for
\begin{align*}
a_0+a_3+a_6+\cdots
\end{align*}
with $(1+x+x^2)^n=\sum_{r=0}^{2n}a_rx^n$ it is convenient to consider the third roots of unity
\begin{align*}
1,\,\omega_1=\frac{-1+i\sqrt{3}}{2},\,\omega_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488771",
"timestamp": "2023-03-29T00:00:00",
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Solve this : $\,\displaystyle{\frac{dy}{dx}} = \cfrac{2xy \,e^{(x/y)^2}}{y^2(1+e^{(x/y)^2})+2x^2e^{(x/y)^2}}$
Solve this : $\,\cfrac{dy}{dx} = \cfrac{2xy \,e^{(x/y)^2}}{y^2(1+e^{(x/y)^2})+2x^2e^{(x/y)^2}}$
I tried to solve it using the homogeneous equation method:
$$y=vx\\ \cfrac{dy}{dx}=v\,+ x\cfrac{dv}{dx}$$
$\impl... | $$\, \cfrac { dy }{ dx } =\cfrac { 2xy\, e^{ (x/y)^{ 2 } } }{ y^{ 2 }(1+e^{ (x/y)^{ 2 } })+2x^{ 2 }e^{ (x/y)^{ 2 } } } \\ \, \cfrac { dy }{ dx } =\cfrac { 2\, e^{ (x/y)^{ 2 } } }{ \frac { y }{ x } (1+e^{ (x/y)^{ 2 } })+2\frac { x }{ y } e^{ (x/y)^{ 2 } } } \\ y=xt\\ \frac { dy }{ dx } =x\frac { dt }{ dx } +t\\ x\frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that $\frac{6n^2+3n}{2n^2-5}$ converges to 3. Prove the following:
$$\lim_{n\to \infty}\frac{6n^2+3n}{2n^2-5}=3$$
Here's my solution:
Consider the function $\frac{6n^2+3n}{2n^2-5}$.
Then $$|\frac{6n^2+3n}{2n^2-5}-3|= |\frac{6n^2+3n}{2n^2-5}-\frac{6n^2+15}{2n^2-5}|=|\frac{3n-15}{2n^2-5}|$$
The upper bound for th... | $$\left|\frac{6n^2+3n}{2n^2-5}-3\right|= \left|\frac{6n^2+3n}{2n^2-5}-\frac{6n^2\color{red}-15}{2n^2-5}\right|=\left|\frac{3n\color{red}+15}{2n^2-5}\right|$$
Hence upper bound for the numerator when $n> 15$, is $|3n+15| \leq 4n.$
A lower bound for the denominator for $n \geq 3$, is $|2n^2-5| \geq n^2$
It follows that $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Weird integral found in textbook I found the following integral in a calculus textbook.
$$\int_{-1}^1\frac{\mathrm d x}{\sqrt{4-x^2}}$$
What is $\mathrm d x$ doing up there and how do I fix this??? I didn't even know it was possible to write an integral like this. Much help needed.
| The $dx$ is a symbol, that tells you by which variable to integrate. But it is often treated like a factor. Different ways to write this integral would be
$$ \int_{-1}^1 \frac{dx}{\sqrt{4-x^2}} = \int_{-1}^1 \frac{1}{\sqrt{4-x^2}}dx = \int_{-1}^1 dx \frac{1}{\sqrt{4-x^2}}. $$
The first variant is often used to save spa... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $n+1$ divides $\binom{an}{bn}$? Suppose that $a>b>0$ be integers. Is it true that for an integer $n>2$ that
$$n+1|\binom{an}{bn}$$
or is there a counter example. Certainly i think the right hand side would reduce to
$$\frac{an(an-1)(an-2)...((a-1)n+1)}{n(n-1)(n-2)...2\cdot 1}$$
But I'm not seeing how this could ... | There is a counterexample.
Take $(n,a,b)=(3,5,1)$.
$\binom{5\times 3}{1\times 3}=455$ is not divisible by $4$.
| {
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"url": "https://math.stackexchange.com/questions/2492260",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Base case: for $n = 1: 4^1 +6\cdot 1 - 10 = 0$ is divisible by 18.
Inductive Assumption: Assume that for for some $k \in \mathbb{N} :4^k +6k-10$
Proving that $4^{k+1}+6(k... | \begin{align}3\cdot 4^k+6 &= 6(2\cdot4^{k-1}+1) \\
&=6(3\cdot4^{k-1}+1-4^{k-1})\\
&=6(3\cdot4^{k-1}+(1-2^{k-1})(1+2^{k-1}))\\
&=6(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))\end{align}
Notice that $2^{k-1}$ is not divisible by $3$, hence either $(2^{k-1}-1)$ is divisible by $3$ or $(2^{k-1}+1)$ is divisible by $3$.
Hence $(... | {
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Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $ Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $
Show that$ a \equiv 1 \pmod{2^4 } \Rightarrow a^{2^{4-2}} \equiv 1 \pmod{2^4} $
Answer:
$ a \equiv 1 \pmod{2^3} \\ \Rightarrow a^2 \equiv 1 \pmod{2^3} \\ \... | Yes, it is correct.
In general, if $a \equiv 1 \pmod m$,then we have $a^n \equiv 1 \pmod m$ for any $n$.
| {
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"answer_count": 2,
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} |
let $f(x) = \left\lbrace\begin{array}{ll} x^2+3x, &x \geq1\\ x^2-3x+6, & x<1 \end{array}\right.$ then find the limit… Let
$f(x) =
\begin{cases}
x^2+3x, &x \geq1\\[2ex]
x^2-3x+6, & x<1
\end{cases}$
Then find the
$$\lim_{h \to0} \frac{f(1)-f(1-h^2)}{h^2}=?$$
My Try :
$$f(1)=4$$
$$f(1-h^2)=(1-h^2)^2-3(1-h^2)+6=h^4-h^2+... | Yes, it's right (with a minor slip, check your algebra). For computing $f(1)$ you have to use the “upper branch”, so
$$
f(1)=1^2+3\cdot 1=4
$$
whereas, since $h^2>0$ when computing the limit, $1-h^2<1$ and $f(1-h^2)$ requires following the “lower branch”, so
\begin{align}
f(1-h^2)
&=(1-h^2)^2-3(1-h^2)+6\\[4px]
&=1-2h^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
show that:for every $x\in S$, there exsit $y,z,w\in S$, such $x=y^2+z^2+w^2$ Define $S=\{a+b\cdot\dfrac{-1+\sqrt{3}i}{2}|a,b\in Z\}$. Show that for every $x\in S$, there exit $y,z,w\in S$,such
$$x=y^2+z^2+w^2$$where $i$ such $i^2=-1$
My attempt:
Since
$$\left(a+b\dfrac{-1+\sqrt{3}i}{2}\right)^2=\left(a^2-ab-\dfrac{1}{... | Let $\omega = e^{\frac{2\pi}{3}i} = \frac{-1 + \sqrt{3}i}{2}$ be the primitive cubic root of unity. It satisfies the identities
$$1 + \omega + \omega^2 = 0\quad\text{ and }\quad \omega^3 = 1$$
For any $x = a + b\omega \in S$, define $r$ according to following table.
$$\begin{array}{|cc:c:l|}
\hline
a & b & r & x + r^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2}... | First solve the real inequalities
$$
x \le \sqrt{x}+\sqrt{x+1}+\sqrt{x+2} < x+1
\tag{1}$$
Solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}$ is numerically $9.8956$ and solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}=x+1$ is numerically $7.9813$. So solution of (1) is
$$
7.9813 < x \le 9.8956
$$
Finally, assume $x$ is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 2
} |
Prove that $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ without using induction. I have to deduce the following formula $$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6},$$ while using the given formula $$\binom{k}{0}+\binom{k+1}{1}+\cdots+\binom{k+r}{r}=\binom{k+r+1}{r}$$
I tried to find values for $k$, such ... | When I was (much) younger I wanted to find the area of a segment of parabola so I needed the sum of the squares and I found the formula by myself in this way.
Inspired by the famous $1+2+\ldots+n=\dfrac{n(n+1)}{2}$
I supposed that the sum of the square could be a third degree polynomial in $n$
$P(n)=an^3+bn^2+cn$
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Show that $\zeta_K(2) = \frac{\pi^4}{48 \sqrt{2}} $, with $K = \mathbb{Q}(\sqrt{2})$ For the real number field $K = \mathbb{Q}(\sqrt{2})$ the ring of integers is $\mathcal{O}_K = \mathbb{Z}[\sqrt{2}] $. We can solve Pell's equation and so there are units $(1 - \sqrt{2})^k$ with $k \in \mathbb{Z}$. One can show that $... | We have that
$$ L(\chi_8,s) = \sum_{n\geq 0}\left[\frac{1}{(8n+1)^s}-\frac{1}{(8n+3)^s}-\frac{1}{(8n+5)^s}+\frac{1}{(8n+7)^s}\right]$$
equals
$$\prod_{p\equiv \pm 1\!\!\pmod{8}}\left(1-\frac{1}{p^s}\right)^{-1}\prod_{p\equiv \pm 3\!\!\pmod{8}}\left(1+\frac{1}{p^s}\right)^{-1} $$
by Euler's product. On the other hand
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A polynomial can not be the product of two polynomials of degree 2 and 3 I want to show that
the polynomial $x^5+x^2-1$ in $\Bbb{Z}/2\Bbb{Z}[x]$ can not be written can a product of a polynomial of degree $2$ and an other one of degree $3$.
For that I wrote $x^5-x^2+1=(x^2+bx+c)(x^3+dx^2+ex+f)=x^5+(b+d)x^4+(c+e+bd)x^3... | Let us show that the polynomial $x^5-x^2+1$ is irreducible in $\Bbb Z[x]$.
We khnow that if a polynomial in reducible in $\Bbb Z[x]$ then it is reducible in $\Bbb{Z}/p\Bbb{Z}[x]$ for any prime number $p$.
Therefore if we show for example that $x^5+x^2+1$ is irreducible in $\Bbb{Z}/2\Bbb{Z}[x]$ then $x^5-x^2+1$ is irre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_{0}^{\pi\over 2}\cos(x)\ln(\tan{x}){\mathrm dx\over 1+\tan(x)}=\int_{0}^{\pi\over 2}\cos(3x)\ln(\tan{x}){\mathrm dx\over 1+\tan(x)}=-\ln{2}?$
$$I=\int_{0}^{\pi\over 2}\cos(x)\ln(\tan{x}){\mathrm dx\over 1+\tan(x)}$$
$$J=\int_{0}^{\pi\over 2}\cos(3x)\ln(\tan{x}){\mathrm dx\over 1+\tan(x)}$$
How to show that $I=J... | As an alternative to the solution provided by Jack D'Aurizio, I will find a value to the integral $I$ by picking up where you left off.
For the first of your integrals $I_1$, if we write
$$I_1 = \int^1_0 \frac{\ln x}{(1 + x) \sqrt{1 + x^2}} \, dx + \int^\infty_1 \frac{\ln x}{(1 + x) \sqrt{1 + x^2}} \, dx,$$
letting $x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the limit of a function . How can I calculate the following limit:
\begin{equation*}
\lim_{x \rightarrow a}
\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}
\end{equation*}
I feel that I should multiply by the conjugate, but which conjugate?
| Note that the above limit doesn't make sense for $x < a$, since the square root in the numerator has a negative argument, but I avoid writing this.
$$
\frac{\sqrt x - \sqrt a + \sqrt{x-a}}{\sqrt{x^2-a^2}} = \frac{1}{\sqrt{x+a}}\left(\frac{\sqrt x - \sqrt a}{\sqrt{x-a}} + 1\right)
$$
Finally, $\lim_{x \to a} \frac{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Product of quadratic residues in terms of primitive root Let $a$ be a primitive root for prime $p(\geq 3)$. Show that the product of all non-zero quadratic residues is
congruent to $a^\frac{p^2−1}{4}$ and that the product of quadratic nonresidues is congruent to $a^\frac{(p−1)^2}{4}$ modulo $p$.
I know that the pro... | Hint: If $a$ is a primitive root mod $p$, $p$ is an odd prime, then all the non-zero quadratic residues mod $p$ are exactly $a^2,a^4,\ldots,a^{p-1}$.
Proof: $a$ is not a quadratic residue because if $a\equiv b^2\pmod{p}$, then by Fermat's little theorem $a^{\frac{p-1}{2}}\equiv b^{p-1}\equiv 1 \pmod {p}$, contradiction... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the simplest way to compute :$\int^{\pi}_0\bigl(\frac{\sin(x)}{5-4\cos(x)}\bigr)^2dx=\frac{\pi}{24}$ I am trying to compute the following integral.
$$\int^{\pi}_0\biggl(\frac{\sin(x)}{5-4\cos(x)}\biggr)^2dx$$
My attempt was to expand the integrand and make use of the standard change of variables $t =\tan x/2$. ... | Let $z=e^{ix}$ and then
\begin{eqnarray}
&&\int^{\pi}_0\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx\\
&=&\frac12\int^{\pi}_{-\pi}\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx\\
&=&\frac12\int_{|z|=1}\bigg(\frac{\frac{z-\frac1z}{2i}}{5-4\frac{z+\frac1z}{2}}\bigg)^2\frac{dz}{iz}\\
&=&\frac12\int_{|z|=1}\bigg(\frac{z^2-1}{2i(5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substi... | Starting from (as you almost wrote)
$$\frac{2x}{(x+2)^2} = \frac{A}{x+2}+\frac{B}{(x+2)^2},$$
rewrite as
$$\frac{2x}{(x+2)^2} = \frac{A(x+2)+B}{(x+2)^2}
= \frac{Ax+(2A+B)}{(x+2)^2}.$$
From this, one sees that $2x = Ax + (2A+B)$, so that $A=2$ and $2A+B=0$ and thus $B=-4$ (as you correctly derived).
An alternative... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2513766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 0
} |
Sum of series $\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{m^2 - n^2}$ I was trying to solve this series and I have an exam in a week. I can't understand how to find its sum although I managed to rework it by transforming $\frac{1}{m^2 - n^2}$ into $\frac{1}{2m}(\frac{1}{m+n} - \frac{1}{n-m})$.
I'm sure there i... | I get
$-\frac{3}{4m^2}
$.
$\begin{array}\\
s(m)
&=\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{m^2 - n^2}\\
&=\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{2m}(\frac{1}{m+n} + \frac{1}{m-n})\\
&= \frac{1}{2m}\sum_{\underset{m \neq n}{n = 1}}^{\infty}(\frac{1}{m+n} + \frac{1}{m-n})\\
&= \frac{1}{2m}(\sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the integral $\int_{0}^{1} f(x)dx$ for $f(x)+f(1-{1\over x})=\arctan x\,,\quad \forall \,x\neq 0$. Suppose that $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$$f(x)+f\left(1-{1\over x}\right)=\arctan x\,,\quad \forall \,x\neq 0$$
Find $$\int_{0}^1 f(x)\,dx$$
My Attempt :
Replace $x$ by $1/x$ in given equation
$$f\... | An Attempt: Let
\begin{equation}
f(x) + f(1 - \frac{1}{x}) = arctan(x) \tag{1}
\end{equation}
Integrating:
\begin{equation}
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx = \int_0^1 arctan(x) dx
\end{equation}
\begin{equation}
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx = \frac{1}{4}(\pi-\ln 4) \tag{2}
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
When $f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$ is an integer $a,b,m,x$ are positive integers.
For which $x>0$ is $f(x)$ an integer?
$$f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$$
I been trying to play with it, I changed it to:
$$\frac{b^2m-a\left(b+x\right)}{a+m\left(b+x\right)}$$
And then I been trying to say:
$$a+m\left(b+x\right)|... | For every even $x>0$ we have that $f(x)$ is not an integer for all $a,b,m$. To see this, chose an even $x$ and take $a=b=m=x$. Then
$$
f(x)=\frac{-x^3}{-2x^2-x}=\frac{x^2}{2x+1}.
$$
Because $x^2$ is even, and $2x+1$ is odd, the fraction cannot be an integer. Similarly, for $x$ odd, take $a=b=x$ and $m=x+1$ to see that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Limit of $f(x)$ when $x$ goes to zero Let $f(x) = \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}$ . Find value of $\lim_{x \to 0} f(x)$ if it exists . I can solve it using L'Hospital's Rule and Taylor series but I'm looking for another way suing trigonometric identities .
| \begin{align}
\dfrac{1-\cos x+\sin x+\tan x}{\sin^2x+x^3}
&= \dfrac{2\sin^2\frac{x}{2}+\sin x\dfrac{1+\cos x}{\cos x}}{\sin^2x+x^3} \\
&= \dfrac{2\dfrac{\sin^2\frac{x}{2}}{x^2}+\dfrac{\sin x}{x}\dfrac{1+\cos x}{x\cos x}}{\dfrac{\sin^2x}{x^2}+x} \\
&\to\dfrac{\dfrac12+1\times\infty}{1+0}=\infty
\end{align}
as $x\to0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Easiest way to show that $\int_0^{2\pi}\frac{\sin x\,dx}{\left|\sin x\right| +\left|\cos x\right|}=0$ What's the easiest way to show that $$\int_0^{2\pi}\frac{\sin x \,dx}{\left|\sin x\right| +\left|\cos x\right|}=0$$
I was thinking about to change the interval of integration and to show that the function is odd.
$\dis... | $$\int_0^{2\pi}\frac{\sin x}{|\sin x| +|\cos x|}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_{\frac{\pi}{2}}^{\pi}\tfrac{\sin x}{\sin x -\cos x}dx+\int_{\pi}^{\frac{3\pi}{2}}\tfrac{\sin x}{-\sin x-\cos x}+\int_{\frac{3\pi}{2}}^{2\pi}\tfrac{\sin x}{-\sin x +\cos x}dx=$$
$$=\int_0^{\frac{\pi}{2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve $\sqrt{n^3+1}+\sqrt{n^3}>10^3$ I was given this task:
"Find $n_0$ so that for all $n > n_0$:
$$\frac{1}{\sqrt{n^3+1}+\sqrt{n^3}}<\ 10^{-3}$$
This should be equal to:
$$\sqrt{n^3+1}+\sqrt{n^3}>10^3$$
But this is where I am already stuck. Wolfram Alpha gave me this solution for n:
$$n\ >\ \frac{9}{100}\left(\frac{3... | $$\sqrt{n^3+1}+\sqrt{n^3}\gt 10^3$$
is equivalent to
$$\sqrt{n^3+1}\gt 10^3-\sqrt{n^3}$$
For $n$ such that $10^3-\sqrt{n^3}\ge 0$, i.e. $n\le 10^2$, the both sides are non-negative, so squaring the both sides gives
$$n^3+1\gt (10^3-\sqrt{n^3})^2,$$
i.e.
$$\sqrt{n^3}\gt \frac{10^6-1}{2\times 10^3}$$
So, we get
$$n\gt \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Condititon for roots of quartic to be real and two be coincident We wish the roots of the following quartic to be real and and distinct, but two roots should be equal. Eg. Roots should be $a,b,c,c$ where $a,b,c$ are Real and distinct.
$$[x^2-2mx-4(m^2+1)][x^2-4x-2m(m^2+1)]$$
We have to find values of $m$ corresponding ... | Let $f(x)=x^2-2mx-4(m^2+1),g(x)=x^2-4x-2m(m^2+1)$.
Also, let $c\in\mathbb R$ be the double root of $f(x)g(x)$.
We have three cases to consider :
Case 1 : $x=c$ is a double root of $f(x)$
Case 2 : $x=c$ is a double root of $g(x)$
Case 3 : $f(c)=g(c)=0$
*
*Case 1 : If $x=c$ is a double root of $f(x)$, then we have to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof explanation: finding the coefficient of $(r+1)$th term in the expansion of $\left(1-6x\right)^{-\frac{1}{2}}$? Here is the answer of the math..
$\displaystyle\left(1-6x\right)^{-\frac{1}{2}}$
$\displaystyle=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)....\left(-\frac{1}{2}-r+1_{ }\right)}{r!}\left(-... |
We obtain
\begin{align*}
&\color{blue}{\frac{\left(-1\right)^r\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)\cdots\left(r-1+\frac{1}{2}\right)}{r!}\left(-1\right)^rx^r2^r3^r}\\
&\qquad=\frac{\left(-1\right)^r\left(\frac{1}{2}\right)\left(\frac{1+2}{2}\right)\left(\frac{1+4}{2}\right)\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
prove that $a^{25}-a$ is divisible by 30 I'm trying to prove that $30|a^{25}-a$.
First of all I said that it equals
$a(a^3-1)(a^3+1)(a^6+1)(a^{12}+1)$, so that it must be divisible by 2.
Now I want to show it can be divided by $3,6$ using Fermat's little theorem.
I need a bit of direction since I'm kind of lost.
| You need divisibility by $2,3,5$.
$a$ and $a^3-1$ are of different parity, so one of them is odd, one is even. Thie product is of course even.
$a^3\equiv a\pmod{3}$, so one of numbers $a^3-1$, $a$, $a^3+1$ is always divisible by $3$.
Now, for $5$.
*
*The case $5|a$ is trivial.
*$a\equiv 1\pmod{5}\implies a^3\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Particular solution of non-homogeneous recurrence relation can somebody help me with my homework, please? I have to solve this:
$a_{n}$ = $-a_{n-1}$ + $12a_{n-2}$ - 10n + 13 + $7.3^{n}$
$a_{0} = 3$, $a_{1}=24$.
I know to solve this (homogeneous equation):
$a_{n}$ = $-a_{n-1}$ + $12a_{n-2}$
$a_{n}$ = $x^{n}$
$x^{n}$ =... | You have:
$$a_n = -a_{n-1} + 12a_{n-2} - 10n + 13 + 7\cdot 3^n$$
$$a_{n+1} = -a_{n} + 12a_{n-1} - 10(n+1) + 13 + 7\cdot 3^{n+1}$$
Subtracting the first from the second equation you get:
$$a_{n+1} = 13a_{n-1} - 12a_{n-2} - 10 + 14 \cdot 3^n$$
Repeat the same trick to get rid of the constant and you have:
$$a_{n+2} = a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How can I calculate $\alpha=\arccos\left(-\frac{1}{4}\right)$ without using a calculator? How can I calculate $\alpha$, without using a calculator?
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
I know $x = -\frac{1}{4} \implies y= \frac{\sqrt{15}}{4}, $ now how can I calculate $$\arccos\left(-\frac{1}{4}\right)... | You can read the binary digits of $\arccos(x)/\pi$ off the signs of $2\cos(2^kx)$, which is an easy to compute sequence defined recursively with $x_{n+1} = x_n^2-2$.
More precisely, you put a $1$ digit when the product of the signs so far is negative, and a $0$ otherwise :
$\begin{matrix}x_0 & -1/2 & - & - \\
x_1 &-7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 11,
"answer_id": 1
} |
If roots of $x^6=p(x)$ are given then choose the correct option If $x_1,x_2,x_3,x_4,x_5,x_6$ are real positive roots of the equation $x^6=p(x)$ where $P(x)$ is a 5 degree polynomial where $\frac{x_1}{2}+\frac{x_2}{3}+\frac{x_3}{4}+\frac{x_4}{9}+\frac{x_5}{8}+\frac{x_6}{27}=1$ and $p(0)=-1$, then choose the correct opti... | $P(0)=-1$ gives $x_1 x_2 x_3 x_4 x_5 x_6 =1$ Now apply AM-GM to $ x_1/2+ x_2/3+ x_3/4+ x_4/9+ x_5/8 + x_6/27 =1$
\begin{eqnarray*}
1 = \frac{x_1}{2 }+ \frac{x_2}{3 }+ \frac{x_3}{ 4}+ \frac{x_4}{9 }+ \frac{x_5}{8 } + \frac{x_6}{27 } \geq \frac{6 \sqrt[6]{x_1 x_2 x_3 x_4 x_5 x_6}}{6} = 1.
\end{eqnarray*}
For this bou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Factorize $x+{1\over 2}x^2+{1\over 3}x^3+ \cdot\cdot$ by $1-x$ I want to derive explicit formula for given recursive relation below:
$$a_{n+1} = (n + 1)a_n + n!$$
for $n ≥ 0$ and $a_0 = 0$.
I had exploited $EGF$, resulting in:
$$g(x)\cdot(1-x) = x+{1\over 2}x^2+{1\over 3}x^3+ \cdot\cdot$$
Thus to derive the explicit f... | We can obtain the numbers $a_n (n\geq 1, a_0=0)$ by calculating the coefficients of the exponential generating function
\begin{align*}
g(x)=a_1x+a_2\frac{x^2}{2!}+a_3\frac{x^3}{3!}+a_4\frac{x^4}{4!}+\cdots
\end{align*}
We obtain
\begin{align*}
\color{blue}{g(x)}&=\left(x+\frac{1}{2}x^2+\frac{1}{3}x^3+\frac{1}{4}x^4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Condition for the inclination of generators through the principal elliptic section of a hyperboloid
Question: Show that the generators through any one of the ends of an equi-conjugate diameter of the principal elliptic section of the hyperboloid $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2} = 1$ are inclined to ea... | Continuing your solution,
$$\begin{align}
\cos\phi &=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{{l_1}^2+{m_1}^2+{n_1}^2}\sqrt{{l_2}^2+{m_2}^2+{n_3}^2}} \\
\implies \cos\phi &=\frac{a^2\sin^2\theta+b^2\cos^2\theta-c^2}{a^2\sin^2\theta+b^2\cos^2\theta+c^2}
\end{align}$$
Putting $\phi = 60^\circ$ and solving, we get
$$a^2sin^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $\langle 3+8\sqrt{2},7\rangle = \langle 3+\sqrt{2}\rangle$ in the ring $\mathbb{Z}[\sqrt{2}]$? How to prove that the ideals $\langle 3+8\sqrt{2},7\rangle$ and $\langle 3+\sqrt{2}\rangle$ are equal in the ring $\mathbb{Z}[\sqrt{2}]$?
I tried using the factors and reducing to the form of the other and v... | Hint: show that each contains the generators of the other.
Solution:
Note that $7=(3+\sqrt{2})(3-\sqrt{2})$, so $7 \in (3+\sqrt{2})$, hence $7\sqrt{2}$, and $3+\sqrt{2}+7\sqrt{2}=3+8\sqrt{2}\in(3+\sqrt{2})$. Thus $(7,3+8\sqrt{2})\subset (3+\sqrt{2})$. Conversely, since $7\in(7,3+8\sqrt{2})$, $3+8\sqrt{2}-7\sqrt{2} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this recurrence relation? $f(1) = a; f(2) = b; f(x) = 2f(x-1)-f(x-2)+2;$ How to solve this recurrence relation?
*
*$f(1) = a;$
*$f(2) = b;$
*$f(x) = 2f(x-1)-f(x-2)+2;$
Where $a$ and $b$ are positive integers
I want to find $f(x)$ representation with $a$ and $b$ only
Also I am sorry, I really not fam... | By combining all the above clues, we have
\begin{equation}
f(n) = (n-1)b - (n-2)a + (n-1)(n-2)
\end{equation}
To verify this, it is clear that $f(1) = a$ and $f(2) = b$.
Also the simple calculation reveals that
\begin{align}
2f(x-1) - f(x-2) +2 &= 2[(x-2)b - (x-3)a + (x-2)(x-3)] \\
&\hspace{0.2cm} ...- [(x-3)b - (x-4)a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding value of $c+d$
At a point $A(1,1)$ on the ellipse , equation of tangent is $y=x.$ If one of the
foci of the ellipse is $(0,-2)$ and coordinate of center of ellipse is $(c,d)$.
Then find value of $c+d$ (given length of major axis is $4\sqrt{10} unit$)
Attempt : assuming one foci is at $S_{1}(0,-2)$ and other i... | The slope of the normal at $(1,1)$ is $-1$ and the normal bisects the angle between $S_1A, S_2A$. Since the slope of $AS_1$ is 3, slope of normal is $-1$, slope of $AS_2$ is $\frac{1}{3}$. The equation of $AS_2$ is $y-1 = \frac{1}{3}(x-1)$. Thus $\beta -1 = \frac{1}{3}(\alpha-1)$. Substitute this in the equation $AS_1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Proving ${\sum_{n=1}^\infty {1\over F_n}} <4$ I'm trying to prove the sum of Fibonacci numbers' reciprocals is less than 4, which is:
$${\sum_{n=1}^\infty {1\over F_n}} <4$$
It makes me confused because the only information I know about Fibonacci numbers that might be useful are its recurrence relation and general form... | This solution uses only simple arithmetic and some recursion.
First we prove that $F_{n+1}-\dfrac{3}{2}F_n\gt 0$ for all $n\ge 4$. Simplify the left hand part:
$$
F_{n+1}-\dfrac{3}{2}F_n = F_{n}+F_{n-1}-\dfrac{3}{2}F_n = F_{n-1}-\dfrac{1}{2}F_n = F_{n-1}-\dfrac{1}{2}(F_{n-1}+F_{n-2}) = \dfrac{1}{2}(F_{n-1}-F_{n-2}) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 8,
"answer_id": 5
} |
Prove that any field $F$ containing $\sqrt{a}+\sqrt{b}$ also contains $\sqrt{a}$ and $\sqrt{b}$.
Prove that any field $F$ containing $\sqrt{a}+\sqrt{b}$ also contains $\sqrt{a}$ and $\sqrt{b}$.
I started by taking $(\sqrt{a}+\sqrt{b})^2=2\sqrt{ab}+a+b$ and want to conclude that $\sqrt{ab}\in F$ but am unsure of this.... | Hint: $\sqrt{a}-\sqrt{b} = (a-b)(\sqrt{a}+\sqrt{b})^{-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.